Simplicial approximation help

Carl R. · science forum beginner Joined: 01 May 2005 Posts: 38

Hi, can you please explain the following solution to the
problem:

This is an example to illustrate that simplicial approximation
may not exist.
Let f: [0,1]->[0,1] be the map given by x->x^2.
Let K=L be the simplicial complex with vertices 0,1/2,1
and edges <0,1/2> and <1/2, 1>.

The solution consists in showing that we can't find
an admissible vertex map phi: V(K) (set of vertices of K)
to V(L) (set of vertices of L) such that the induced
continuous map |phi|: |K| -> |L| where |K| denotes the
underlying space of K.

So clearly f(0)=0^2=0 and hence carrier(f(0))=carrier(0)
=<0>. And hence since |phi| is an approximation of f
then |phi(0)| belongs to carrier(f(0))=<0> and hence
|phi(0)|=0.

Ok so far this is clear. The next part is the one I don't get.
Then the author says:

Since |phi(0)|=0 this implies phi(0)=0. Why is this???
I understand that since <0> is a simplex of K then by
definition of admissible vertex map <phi(0)> must be mapped to
a vertex of L, and the only vertices of L are
0,1/2 and 1, hence phi(0)=0 or 1/2 or 1. But why the author
concludes phi(0)=0 ?

Then it says , f(1)=1^2 and hence phi(1)=1.
Again I don't get why necessarily this implies phi(1)=1.

Then:
Since phi is an admissible vertex map we must have phi(1/2)=1/2.
How do you know that 1/2 must get mapped to 1/2 and not other point?

Then it says for x=2/3 , f(x)=4/9 with carrier <0,1/2>.
So the carrier condition fails for this point since
|phi|(2/3)= 2/3 which is not in <0,1/2>.

Again my question why is |phi|(2/3) = 2/3?

Thanks in advance.

John H Palmieri · science forum beginner Joined: 04 May 2005 Posts: 13

On May 04 2005, "Carl R." <solrac140@hotmail.com> wrote:

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